Thread: Trig Limit Help

1. Trig Limit Help

1. limx->0 (cos(x)-1)/sin(2x)

I have it down to (-sinx^2)/((sin2x)(1+cosx)), but I can't seem to finish it

2. For my second question I don't even have a start. Here it is.

limx->0 (ln(1+x)/(sin3x))

Thank you so much in advance

2. Originally Posted by chrisski
1. limx->0 (cos(x)-1)/sin(2x)

I have it down to (-sinx^2)/((sin2x)(1+cosx)), but I can't seem to finish it
I'm not sure how you arrived at this but maybe this will help.

$\frac{-\sin^2(x)}{(\sin(2x))(1+\cos(x))}$

$\frac{-\sin^2(x)}{(2\sin(x)\cos(x))(1+\cos(x))}$

$\frac{-\sin(x)}{(2\cos(x))(1+\cos(x))}$

3. I could very well have arrived at an incorrect point. If I'm doing it incorrectly please inform me. Also, any help on the second one would be appreciated.

4. Originally Posted by chrisski
I could very well have arrived at an incorrect point. If I'm doing it incorrectly please inform me. Also, any help on the second one would be appreciated.
$\lim_{x \to 0} \frac{ln(1+x)}{\sin 3x}=\lim_{x \to 0} \left ( \frac{ln(1+x)}{\sin 3x} \cdot \frac{3x}{3x} \right )$
$= \lim_{x \to 0} \left ( \frac{ln(1+x)}{ 3x} \cdot \frac{3x}{\sin 3x} \right )$
$=\frac{1}{3} \lim_{x \to 0} \left (\frac{ln(1+x)}{ x} \right ) \cdot \lim_{x \to 0} \left (\frac{1}{\frac{\sin 3x}{3x}} \right )$
$=\frac{1}{3} \cdot 1 \cdot 1 = \frac{1}{3}$

5. Originally Posted by pickslides
I'm not sure how you arrived at this but maybe this will help.

$\frac{-\sin^2(x)}{(\sin(2x))(1+\cos(x))}$

$\frac{-\sin^2(x)}{(2\sin(x)\cos(x))(1+\cos(x))}$

$\frac{-\sin(x)}{(2\cos(x))(1+\cos(x))}$
$\frac{cosx-1}{sin2x}=\frac{cosx-1}{sin2x}\cdot \frac{cosx+1}{cosx+1} = \frac{(cosx-1)(cosx+1)}{sin2x(cosx+1)} =$ $\frac{cos^2(x)-1}{sin2x(cosx+1)} = \frac{-sin^2(x)}{sin2x(cosx+1)}$